I got the following question:
Let $S$ be the total claim amount. Find $P(S=s) ,s= 0,...,6$, if $S = Y_1 + 2Y_2 + 3Y_3, Y_ j∼Poisson(j)$.
I don't get why they first add $Y_1$ and $Y_2$ and then $Y_3$. Multiplying column 2 and 3 doesn't result in column 4, it gives different numbers. And also, why is the distribution of $2Y_2$ given by $\frac{2^{(x/2)}}{(x/2)!}$ (and similar for $3Y_3$)? I included the link of the answer below:
Think in t his way:
$Y\sim Po(2)$
find the distribution of
$Z=2Y$
By definition you get
$$P(Z=z)=P(2Y=z)=P(Y=z/2)=\frac{e^{-2}2^{z/2}}{(z/2)!}$$
This is the distribution of $2Y_2$
similarly for $3Y_3$
Now, as an example, to calculate
$$P(S=0)=P(Y_1=0,Y_2=0;Y_3=0)=e^{-6}$$
$$P(S=2)=P(Y_1=2;Y_2=0;Y_3=0)+P(Y_1=0;Y_2=1;Y_3=0)=e^{-6}[2+\frac{1}{2}]$$
same results as in your notes....is it clear how to proceed?
Let's focus for example on $S=3$, as you stated in your comment:
the result is $P(S=3)=e^{-6}[5+\frac{1}{6}]$ and it is given by summing the numbers in the blue circles: $2+\frac{1}{6}+3$ and multyplied by $e^{-6}$ above in the first row.
this can be achieved without the distribution of $2Y_2$ and $3Y_3$ they calculated but simply summing all the interested events.
For example, to get $S=3$ you can have
$\{y_1,y_2,y_3\}=$
$\{3,0,0\} \xrightarrow{\text{with probability}}e^{-6}[\frac{1}{3!}\times1 \times 1]$
$\{1,1,0\} \xrightarrow{\text{with probability}}e^{-6}[1\times2 \times 1]$
$\{0,0,1\} \xrightarrow{\text{with probability}}e^{-6}[1\times1 \times 3]$
Concluding
$$P(S=3)=e^{-6}[\frac{1}{6}+2+3]=e^{-6}\times 5\frac{1}{6}$$
using the "strange" notation of your notes.